Question

m为有理数，问K为何值时，方程x^2-4mx+4x+3m^2-2m+4k=0的根为有理数。写过程。谢谢

Answer 1

x^2-4mx+4x+3m^2-2m+4k=0
=>x+(4-4m)x+(3m^2-2m+4k)=0
方程要有根，则
Δ=（4-4m)^2-4(3m^2-2m+4k)>=0
=>4m^2-24m+(16-16k)>=0
=>m^2-6m+4(1-k)>=0
同时对任何m都需要m^2-6m+4(1-k)必须为平方数,推出k=-5/4

Answer 2

x^2-4mx+4x+3m^2-2m+4k=0
=>x^2+(4-4m)x+(3m^2-2m+4k)=0
方程要有根，则
Δ=（4-4m)^2-4(3m^2-2m+4k)>=0
=>4m^2-24m+(16-16k)>=0
=>m^2-6m+4(1-k)>=0=>（m-3)^2>=4k+5
所以 4k+5<=0

k<=-5/4

Answer 3

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